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Question
Noah noticed that an air conditioner from Company A can cool a hall in 10 minutes, while an air conditioner from Company B can do the same in 15 miutes.
It is a very hot day, so Noah has turned on both the air conditioners.
Find how long both the air conditioners will take to cool the hall.
Noah noticed that an air conditioner from Company A can cool a hall in 10 minutes, while an air conditioner from Company B can do the same in 15 miutes.
It is a very hot day, so Noah has turned on both the air conditioners.
Find how long both the air conditioners will take to cool the hall.
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Solution
\(Company~ A ~air~conditioner:\)
Time taken to cool the hall = 10 minutes
Work done in 1 minute = \(\dfrac{1}{10}\)
\(Company~B~air~conditioner:\)
Time taken to cool the hall = 15 minutes
Work done in 1 minute = \(\dfrac{1}{15}\)
\(Both~air~ conditioners:\)
Total work done in 1 minute \(= \dfrac{1}{10} +\dfrac{1}{15}\)
= \(\dfrac{15~+~10}{15~\times~10}\)
\(= \dfrac{25}{150}\)
\(= \dfrac{1}{6}\)
The time taken to cool the hall will be the inverse of the work done in 1 minute.
Thus,
Time taken to cool the hall = 6 minutes
Both the air conditioners together will take 6 minutes to cool the hall.
Time taken to cool the hall = 10 minutes
Work done in 1 minute = \(\dfrac{1}{10}\)
\(Company~B~air~conditioner:\)
Time taken to cool the hall = 15 minutes
Work done in 1 minute = \(\dfrac{1}{15}\)
\(Both~air~ conditioners:\)
Total work done in 1 minute \(= \dfrac{1}{10} +\dfrac{1}{15}\)
= \(\dfrac{15~+~10}{15~\times~10}\)
\(= \dfrac{25}{150}\)
\(= \dfrac{1}{6}\)
The time taken to cool the hall will be the inverse of the work done in 1 minute.
Thus,
Time taken to cool the hall = 6 minutes
Both the air conditioners together will take 6 minutes to cool the hall.
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